F(x)=3/2x+9

Solution for F(x)=3/2x+9 equation:

(F)=3/2F+9

We move all terms to the left:

(F)-(3/2F+9)=0


Domain of the equation: 2F+9)!=0

F∈R


We get rid of parentheses

F-3/2F-9=0

We multiply all the terms by the denominator


F*2F-9*2F-3=0

Wy multiply elements

2F^2-18F-3=0

a = 2; b = -18; c = -3;
Δ = b2-4ac
Δ = -182-4·2·(-3)
Δ = 348
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{87}}{2*2}=\frac{18-2\sqrt{87}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{87}}{2*2}=\frac{18+2\sqrt{87}}{4}
$